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You are trying to lose weight by working out on a rowing machine. Each time you pull the rowing bar (which simulates the "oars") toward you, it moves a distance of \(1.2 \mathrm{~m}\) in a time of \(1.5 \mathrm{~s}\). The readout on the display indicates that the average power you are producing is \(82 \mathrm{~W}\). What is the magnitude of the force that you exert on the handle?

Short Answer

Expert verified
The force exerted is approximately 102.5 N.

Step by step solution

01

Understand the Power Relationship

Power is defined as the rate at which work is done or energy is transferred. Mathematically, power can be expressed as \( P = \frac{W}{t} \), where \( P \) is power in watts, \( W \) is work done in joules, and \( t \) is time in seconds.
02

Define Work in Terms of Force and Distance

Work (\( W \)) done by a force is defined as \( W = F \cdot d \), where \( F \) is the force applied in newtons, and \( d \) is the distance moved in the direction of the force in meters. In this case, the distance \( d \) is \( 1.2 \mathrm{~m} \).
03

Express Power in Terms of Force and Distance

Combine the formulas from the previous steps: \( P = \frac{W}{t} = \frac{F \, \cdot \, d }{t} \). We know \( P = 82 \mathrm{~W} \), \( d = 1.2 \mathrm{~m} \), and \( t = 1.5 \mathrm{~s} \).
04

Calculate Force Using Known Values

Rearrange the equation to solve for force: \( F = \frac{P \cdot t}{d} \). Substitute the known values into the equation: \( F = \frac{82 \cdot 1.5}{1.2} \).
05

Solve for the Magnitude of Force

Calculate \( F \): \( F = \frac{82 \times 1.5}{1.2} = 102.5 \mathrm{~N} \). Therefore, the magnitude of the force exerted on the handle is approximately \( 102.5 \mathrm{~N} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power
When you think of power, imagine how quickly you can do a certain amount of work. Power is essentially a measurement of how fast or slow you are performing an action. It's defined as the rate at which work is completed or energy is transferred over time.
Mathematically, power is expressed as \( P = \frac{W}{t} \), where \( P \) represents power in watts, \( W \) is the work done in joules, and \( t \) is the time in seconds. This formula helps us understand that the more power we need to do a task, the less time it should take, assuming work done stays constant.
Relating back to our rowing machine example, if you’re pulling hard and fast, you’re exerting more power as compared to pulling slowly and gently.
Work-Energy Principle
Work is closely related to energy — they can be thought of as best friends in physics. When you do work, you are essentially transferring energy to an object. The work done on an object by a force is equal to the change in energy of that object.
In our rowing machine example, when you pull the oar, you exert a force over a distance, which is doing work. According to the work-energy principle, this work done on the oar is transforming your energy into a motion of the handle, causing a displacement over time.
Remember, the work done is calculated using the formula: \( W = F \cdot d \), where \( W \) is the work, \( F \) the force applied, and \( d \) the displacement/distance moved by the force. This conversion of energy makes the activity of rowing incredibly efficient for workouts.
Distance and Force Relationship
Distance and force have an important relationship when calculating work done. Essentially, to calculate work, you need to understand how far an object is moved and how much force is applied. The greater the distance and the higher the force, the more work done.
In the example of the rowing machine, each stroke moves the handle 1.2 meters, and your goal is to find out the force you apply. The power output is given as 82 watts, which helps calculate the force when understanding how much energy you're generating each time you row.
By using the formula for work \( W = F \cdot d \), and understanding the definition of power, you can derive the formula: \( P = \frac{F \cdot d}{t} \). Then you rearrange it to solve for force: \( F = \frac{P \cdot t}{d} \). This demonstrates how distance traveled affects the amount of force needed and consequently the work done.
Physics Problem Solving
Solving physics problems often feels like piecing together a puzzle. You need to understand each part to see the full picture. Imagine having several tools (like formulas) that you need to use correctly to find a solution.
Let's break down our rowing machine problem: you know the handle moves 1.2 meters in 1.5 seconds with an average power of 82 watts. Using the relationship of power and the work-energy principle, you can find the force applied.
First, rearrange your equation \( P = \frac{F \cdot d}{t} \) to find \( F \) (force): \( F = \frac{P \cdot t}{d} \). Fill in the known numbers to solve for \( F \) and you get approximately 102.5 newtons. The key is to ensure you carefully follow the logical steps and understand how each element fits together. Physics problem-solving is about applying the right equations and understanding where and how they work best.

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Most popular questions from this chapter

A shot-putter puts a shot (weight \(=71.1 \mathrm{~N}\) ) that leaves his hand at a distance of \(1.52 \mathrm{~m}\) above the ground. (a) Find the work done by the gravitational force when the shot has risen to a height of \(2.13 \mathrm{~m}\) above the ground. Include the correct sign for the work. (b) Determine the change \(\left(\Delta \mathrm{PE}=\mathrm{PE}_{\mathrm{f}}-\mathrm{PE}_{0}\right)\) in the gravitational potential energy of the shot

Relative to the ground, what is the gravitational potential energy of a 55.0 -kg person who is at the top of the Sears Tower, a height of \(443 \mathrm{~m}\) above the ground?

A \(55.0-\mathrm{kg}\) skateboarder starts out with a speed of \(1.80 \mathrm{~m} / \mathrm{s} .\) He does \(+80.0 \mathrm{~J}\) of work on himself by pushing with his feet against the ground. In addition, friction does \(-265 \mathrm{~J}\) of work on him. In both cases, the forces doing the work are nonconservative. The final speed of the skateboarder is \(6.00 \mathrm{~m} / \mathrm{s}\). (a) Calculate the change \(\left(\Delta \mathrm{PE}=\mathrm{PE}_{\mathrm{f}}-\mathrm{PE}_{0}\right)\) in the gravitational potential energy. (b) How much has the vertical height of the skater changed, and is the skater above or below the starting point?

A pitcher throws a \(0.140-\mathrm{kg}\) baseball, and it approaches the bat at a speed of \(40.0 \mathrm{~m} / \mathrm{s}\). The bat does \(W_{n c}=70.0 \mathrm{~J}\) of work on the ball in hitting it. Ignoring air resistance, determine the speed of the ball after the ball leaves the bat and is \(25.0 \mathrm{~m}\) above the point of impact.

Two cars, \(A\) and \(B\), are traveling with the same speed of \(40.0 \mathrm{~m} / \mathrm{s}\), each having started from rest. Car A has a mass of \(1.20 \times 10^{3} \mathrm{~kg}\), and car \(\mathrm{B}\) has a mass of \(2.00 \times 10^{3} \mathrm{~kg} .\) Compared to the work required to bring car A up to speed, how much additional work is required to bring car B up to speed?

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